If $\displaystyle F(x)= sqrt(x) $, describe the transformation from $\displaystyle F(x) $to $\displaystyle G(x)$ if $\displaystyle G(x) = -f(-x+1)-7 $
If, alternatively, you are looking for a description of the graph of the new function (i.e. shifts/rotations about lines/stretching/skewing/etc.), then break it down step-by-step.
Compare the graph of $\displaystyle f(x)$ to the graph of $\displaystyle f(x-1)$. Then compare those to the graph of $\displaystyle f(-(x-1)) = f(-x+1)$. Next, compare them to the graph of $\displaystyle -f(-x+1)$. Finally, put it all together as you compare it to the graph of $\displaystyle -f(-x+1)-7$.
Let's consider what happens when we plug in numbers. If we plug in 5 to $\displaystyle f(x-1)$, we get $\displaystyle f(5-1) = f(4)$. Going from the graph of $\displaystyle f(x)$ to the graph of $\displaystyle f(x-1)$, the x-position of $\displaystyle f(4)$ is going from x=4 to x=5. In other words, it is shifting to the right one unit. Now let's consider $\displaystyle f(-(x-1))$. Now, $\displaystyle f(-4)$ is at x=-4 in our original graph, but it is at x=5 in this graph. Does this mean it is shifting to the right nine units? Not quite. Let's look at some more data to see what's happening. $\displaystyle f(-3)$ is the y-value at x=-3 in our original graph, but it is the y-value at x=4 in $\displaystyle f(-(x-1))$. So now the shift is only 7 units. The negative sign is actually creating a reflection across some axis. The closer we get to $\displaystyle x=1$, the less the points need to shift from the original function to our new function. In other words, this is a reflection across the line x=1. Next, $\displaystyle -f(-x+1)$ is taking the value $\displaystyle f(-x+1)$ and taking its reflection across the x-axis. Finally, $\displaystyle -f(-x+1)-7$ is taking the graph of $\displaystyle -f(-x+1)$ and shifting it down seven units.